Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline char nc() {
  static char buf[100000], *p1, *p2;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <class T>
inline void rd(T& x) {
  x = 0;
  char ch = nc();
  for (; !isdigit(ch); ch = nc())
    ;
  for (; isdigit(ch); ch = nc()) x = x * 10 - 48 + ch;
}
const int maxn = 1e5 + 10;
int nx[maxn << 1], to[maxn << 1], hd[maxn], ek = 1;
inline void adde(int u, int v) {
  to[ek] = v;
  nx[ek] = hd[u];
  hd[u] = ek++;
}
vector<int> v[maxn];
int sum[maxn * 40], lson[maxn * 40], rson[maxn * 40], nsize, root[maxn];
void update(int pos, int pre, int& now, int l, int r) {
  now = ++nsize;
  lson[now] = lson[pre];
  rson[now] = rson[pre];
  sum[now] = sum[pre];
  ++sum[now];
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update(pos, lson[pre], lson[now], l, mid);
  else
    update(pos, rson[pre], rson[now], mid + 1, r);
}
int m;
int size[maxn], top[maxn], dep[maxn], fa[maxn];
void dfs(int u) {
  size[u] = 1;
  top[u] = u;
  int son = 0;
  root[u] = root[fa[u]];
  for (int i = 0; i < v[u].size(); i++) {
    int x = v[u][i];
    update(x, root[u], root[u], 1, m);
  }
  for (int i = hd[u]; i; i = nx[i]) {
    int v = to[i];
    if (v == fa[u]) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    dfs(v);
    size[u] += size[v];
    if (size[v] > size[son]) son = v;
  }
  if (son) top[son] = u;
}
int query(int a, int b, int c, int d, int k, int l, int r) {
  int s = lson[a][sum] + lson[b][sum] - lson[c][sum] - lson[d][sum];
  if (l == r) {
    if (k == 1 && sum[a] + sum[b] - sum[c] - sum[d] == 1) return l;
    return 0;
  }
  int mid = l + r >> 1;
  if (k <= s) return query(lson[a], lson[b], lson[c], lson[d], k, l, mid);
  return query(rson[a], rson[b], rson[c], rson[d], k - s, mid + 1, r);
}
int findtop(int u) { return u == top[u] ? u : top[u] = findtop(top[u]); }
int lca(int u, int v) {
  if (findtop(u) != findtop(v))
    return top[u][dep] > top[v][dep] ? lca(top[u][fa], v) : lca(u, top[v][fa]);
  return dep[u] > dep[v] ? v : u;
}
int main() {
  int n, q;
  rd(n);
  rd(m);
  rd(q);
  for (int i = 1; i < n; i++) {
    int u, v;
    rd(u);
    rd(v);
    adde(u, v);
    adde(v, u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    rd(x);
    v[x].push_back(i);
  }
  dfs(1);
  vector<int> ans;
  for (int i = 1; i <= q; i++) {
    int u, v, k;
    ans.clear();
    rd(u);
    rd(v);
    rd(k);
    int l = lca(u, v), lf = fa[l];
    for (int j = 1; j <= k; j++) {
      int z = query(root[u], root[v], root[l], root[lf], j, 1, m);
      if (z)
        ans.push_back(z);
      else
        break;
    }
    printf("%d ", ans.size());
    for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}